Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $r = \dfrac{n^2 - 10n}{n + 9} \div \dfrac{n^2 - 18n + 80}{2n + 18} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{n^2 - 10n}{n + 9} \times \dfrac{2n + 18}{n^2 - 18n + 80} $ First factor the quadratic. $r = \dfrac{n^2 - 10n}{n + 9} \times \dfrac{2n + 18}{(n - 10)(n - 8)} $ Then factor out any other terms. $r = \dfrac{n(n - 10)}{n + 9} \times \dfrac{2(n + 9)}{(n - 10)(n - 8)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ n(n - 10) \times 2(n + 9) } { (n + 9) \times (n - 10)(n - 8) } $ $r = \dfrac{ 2n(n - 10)(n + 9)}{ (n + 9)(n - 10)(n - 8)} $ Notice that $(n + 9)$ and $(n - 10)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ 2n\cancel{(n - 10)}(n + 9)}{ (n + 9)\cancel{(n - 10)}(n - 8)} $ We are dividing by $n - 10$ , so $n - 10 \neq 0$ Therefore, $n \neq 10$ $r = \dfrac{ 2n\cancel{(n - 10)}\cancel{(n + 9)}}{ \cancel{(n + 9)}\cancel{(n - 10)}(n - 8)} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $r = \dfrac{2n}{n - 8} ; \space n \neq 10 ; \space n \neq -9 $